Voiceover: Here's another
carboxylic acid derivative so, this is an acid anhydride
over here on the right. And we can form those
from carboxylic acids. So if we start with the carboxylic acid, and our first step, add a base, like sodium hydroxide,
and our second step, add an acyl chloride, then we'd form our acid anhydride as our product. Now if you think about a mechanism, sodium hydroxide's a base. The hydroxide anion is
gonna take this proton, leaving these electrons
behind on the oxygen. There are already two lone pairs of electrons on the oxygen to start with, so if we go ahead and draw the product, we would form a carboxylate anion. So, this oxygen right here would have three lone pairs of
electrons on it, like that. And so if we follow
some of those electrons, this would have a -1 formal charge, and if we put these electrons in magenta, those electrons come off onto our oxygen to form our carboxylate anion. And that's gonna function
as our nucleophile. So in the second step,
we add our acyl chloride. I'm just gonna go ahead and draw in our acyl chloride. It's going to be our electrophile, and let's think about why here. We have the oxygen
withdrawing electron density from this carbon, because oxygen is more electronegative than
carbon, so we have that. And then we also have this
chlorine doing it as well. Chlorine is more electronegative
than carbon as well, so we have these two things withdrawing electron density, and so this carbon is definitely electrophilic, right here. And so we have a
nucleophile that's going to attack our electrophile,
so our nucleophile attacks our electrophile, these electrons kick off onto our oxygen,
and we can go ahead and show the result of our nucleophilic attack. So we would now have this carbon double-bonded to this
oxygen, and then this oxygen is now bonded to
this carbon, and then we have an oxygen up here, with three lone pairs of electrons, -1 formal charge. We still have this carbon bonded to a chlorine, and we still have an R prime group here, like that. So following some electrons,
let's go ahead and put in these lone pairs here,
the electrons in magenta form the bond between the
oxygen and the carbon, and then we could say that
these electrons in here moved off onto our oxygen, to give that oxygen a -1 formal charge. When we think about the
next step, we know that the chloride anion is an
excellent leaving group. So, if these electrons
in blue move in here to reform our double
bond, then these electrons would kick off onto the chlorine to form the chloride anion,
which we know is a stable on its own, as a leaving group. And so, that forms our acid anhydride, so let's go ahead and draw in
the final structure here. So we would have our oxygen, with lone pairs of electrons on our oxygen. We would have this bond. We reformed our carbonyl like that, and then the chloride anion was our leaving group, so now we
have an R prime group, and so we've formed our acid anhydride. And so we could make
these R groups the same, or we could make them
different, and so this is a good way of forming a mixed anhydride, as well as one that is symmetrical. So let's look at an example. So let's take acetic
acid and we're gonna do two different reactions with acetic acid. So the first thing we're gonna do, is add thionyl chloride to acetic acid, and we've seen that the
addition of thionyl chloride converts a carboxylic
acid into an acyl halide. So let's go ahead and show the conversion of that carboxylic acid into the corresponding acyl halide. And then we could take that acetic acid, and in a separate reaction, we could add sodium hydroxide, and
the hydroxide takes this proton, which leaves
these electrons behind on the oxygen. So let's scoot up a little bit, so we can see the mechanism that
we talked about before. And so this would form sodium acetate, draw in our lone pairs of electrons on this oxygen, -1 formal
charge, and a plus like that. So this oxygen already had two lone pairs of electrons. And so now you have the situation that we talked about up here. Your carboxylate anion
functions as a nucleophile, attacks your electrophilic carbon on your acyl chloride. So we could show these electrons attacking this carbon, these electrons
kick off onto the oxygen, and then when those
electrons move back in, to reform your double
bond, these electrons would kick off onto your chlorine, and then so we can go ahead and draw our product. So just thinking about
what happens in this mechanism, we can go ahead and draw our products, which would be
a symmetical anhydride, so we would have our oxygen right here, and then we would have
our group over here. So thinking about the R groups this time, so this R group is a
methyl group, and then we could think about this oxygen being this oxygen, and then this
portion of the acyl chloride is this portion for our final product for our acid anhydride. And so this is acetic
anhydride, which is the one that's used most commonly
in an undergraduate lab. And so this is a nice way of preparing acid anhydrides. Let's look at another way
to form acetic anhydride. You could start with two carboxylic acids, and this would be acetic
acid and acetic acid, so the same carboxylic acid, and apply high heat, and this time you think about a dehydration reaction,
so you could think about losing an OH from one, and a hydrogen from the other, to form
water, so you could think about losing a water here, and your dehydration reaction. And then you can stick
those portions of the molecule together, so you
could take this portion, and then this portion,
and put them together, and you can see that that is once again acetic anhydride, so let's
go ahead and draw that. So we would form acetic anhydride here by dehydration. So this way of doing it is
not always the best way, it works for acetic
acid, but it doesn't work for most carboxylic acids. Here's one more case where it can work, if you have a situation like this. This is phthalic acid, so
it's a dicarboxylic acid, And if you apply heat
to it, you don't need as high of a heat as you need
for the previous reaction, this heat is higher than this one, but you can once again form an anhydride. So if you think about
a dehydration, losing OH from one, and H from the other, and then we can go ahead
and draw the product here, so we would form our benzene ring, and then we would form
our anhydride like this. So once again, loss of water. So the name of this
acid anhydride would be phthalic anhydride because it's derived from phthalic acid. So this is a good way to form five- or six-membered rings. In this case, we have
a five-membered ring. We have a carbon, an oxygen, a carbon, a carbon, and a carbon, so we have a five-membered ring this time. It also works for six-membered rings.